Intensity of EM Waves - Online Tutor, Practice Problems & Exam Prep

All waves, including mechanical, sound, and electromagnetic waves, carry energy, which can be measured as intensity (I). Intensity is defined as power (P) per area (A), expressed as I^2 = PA. For electromagnetic waves, intensity relates to electric field (E) and magnetic field (B) strengths through equations like I = 1/2cEmax. Understanding these relationships is crucial for solving problems involving light sources and their intensity at various distances.

Intensity of EM Waves

Hey, folks. Welcome back. One of the things we talked about with mechanical waves is that they carry energy, whether it was waves on a string, an ocean wave, or something like that. But that's not a property that's unique to mechanical waves. Remember that all waves carry energy. We've got sound waves. We've got light that's coming from the sun. All of those things carry energy. But a more useful measure isn't energy, but it was a related unit or measurement called intensity. Now I'm going to spend some time, but remember that intensity was basically defined as the energy per time divided by area, or you might remember that also as the power per area. Remember, energy over time is just power. We saw this equation here that intensity was just equal to p/A, and that's still true for electromagnetic waves. But now we're going to see some new equations here that relate the intensity of electromagnetic waves to the magnitudes of the electric and magnetic fields here. So there's a couple of new equations that you're going to see, and I'm just going to go ahead and list them all for you. The first one is that the intensity \( I \) is equal to \( \frac{1}{2} c \epsilon_0 \times E_\text{max}^2 \). Now you might see this written in a bunch of different ways in your textbooks. They like to throw out a bunch of different variations of this equation, but the other one that you'll see probably pretty commonly is \( \frac{1}{2} \frac{c}{\mu_0} \times B_\text{max}^2 \). If you haven't seen this, I highly recommend that you learn it this way because I think it's really easy. One of the things that's really easy to see about these equations is that both of them have a factor of \( \frac{1}{2} c \), and then basically both of them have some kind of maximum value squared, whether it's \( E \) or \( B \). And the last thing here is that \( \epsilon \) usually goes with \( E \) and \( \mu \) usually goes with \( B \). So I think this is a really great way to memorize these equations. It also sort of looks like other energy-like equations where you have \( \frac{1}{2} \) times something times something squared.

The units, just as we learned when we talked about mechanical waves, are going to be watts per meter squared. And basically, there are 2 types of problems when it comes down to this intensity stuff. There's one problem in which the source of light will emit radially in all directions like a flashlight or like a bulb or like a star or something like that. Or you might have something where it emits directionally, like a flashlight or a laser or something like that. And basically, the gist of these two different types of problems is that if you can assume the source emits equally in all directions, basically if you can assume that it's this case, then that means that the area, this \( A \) term that pops up in your area equation in the intensity, is going to be equal to \( 4\pi r^2 \) because, basically, if you draw some little distance \( r \), then the surface area of this field basically just becomes the surface area of a sphere. However, if it emits directionally like a flashlight, then you cannot assume that this area is equal to \( 4\pi r^2 \), and you're going to have to figure it out. That's basically the hardest parts of these kinds of problems is figuring out which one you're dealing with.

Now I'm going to go ahead and make a point here about the second thing, but, actually, before I do that, let's just jump right into a problem. Alright? So we have an incandescent light bulb that is emitting at 50 watts, in all directions. So because it's all directions here, that means that we can actually go ahead and use this relationship here. So that means we know what our area is going to be. It's got no inefficiencies or energy losses, and we want to calculate what is the intensity of light at some distance away from the light bulb. Okay? So in this first part here, I want to calculate \( I \). That's basically what I'm looking at here. Okay? So remember that \( I \) has a couple of different formulas. It's got power over area, and it's got all these other things with the \( E \)'s and the \( B \)'s. Which one do I use? Well, if you take a look here, this unit, this 50 watts is actually the power of the light bulb. So that's in my \( P \), so I've got that. And then we know that the light bulb emits in all directions, so we actually already know what the area is as well. So we actually have both of these, and we can go ahead and calculate this. Alright? So if you take a look, let's just say that we're going to draw this distance and we're going to say that \( r = 5 \) meters. And, basically, the area, the surface area here, just becomes, \( 4\pi r^2 \). Alright? So we're going to have that power is equal to 50 divided by the area, which is going to be \( 4\pi \times 5^2 \). You've got to work this out. What you're going to get is \( 0.16 \) watts per square meter, and that is your final answer for part a.

Now we're going to move on to the second part here, which is we're going to calculate the maximum value of the electric field at this distance here. Alright? So we're going to calculate the electric field. So we'll go back to our equations for intensity. Remember that intensity is related to the magnitudes or the strengths of the electric and magnetic fields, and we're going to see this equation here. So we're going to have \( \frac{1}{2} c \epsilon_0 E_\text{max}^2 \). Since we're looking for the electric field, we're going to use this version instead of the \( B_\text{max} \) version. Alright? So, basically, what I'm going to do is I'm going to say that \( I \) is equal to, this is still equal to \( P/A \), but now it's equal to \( \frac{1}{2} c \epsilon_0 \times E_\text{max}^2 \). Now I'm really looking for what is this \( E_\text{max} \) over here, so I can just rearrange my equation. Alright? So I'm not going to use this \( P/A \) because I actually already figured out what this \( I \) is equal to. Right? It's just equal to this number over here. Alright? So this is just equal to 0.16, and this is going to equal, let's see. I'm going to move all this other stuff over to the other side. When I move this to this side, basically, what happens is I pick up a factor of 2 on the outside over here, and then I have to divide by \( c \) and \( \epsilon_0 \). So this is equal to my \( E_\text{max}^2 \). Now the last thing I have to do here is I just have to take the square root and then plug in all of my numbers. So in other words, \( E_\text{max} \) is equal to the square root of \( 2 \times 0.16 \). And now I'm going to plug in my numbers. This is going to be \( 3 \times 10^8 \). And then \( \epsilon_0 \), that's just a constant that we've got over here in this table right here. So we've got 8.85 times \( 10^{-12} \). When you go ahead and plug all of this into your square root, what you're going to get here, is you should get an \( E_\text{max} \) of 10.98. Remember that the units for electric field are newtons per coulomb. So that is the answer to part b. Let's move on now to our final piece here, our final step, which is that we're going to calculate now the RMS value of the magnetic field at this point. And that actually brings me to the second point that I had in this video, which is that in some problems, you might be given or asked for some kind of an average value or some kind of a root mean square. Remember we used that in thermodynamics a little bit, instead of maximum values. And if that ever happens, there's actually a pretty straightforward relationship between RMS and max values, And it's just that, RMS is equal to the maximum divided by the square root of 2. So your \( E_\text{RMS} \) is going to be \( E_\text{max} / \sqrt{2} \). Your \( B_\text{RMS} \) is going to be \( B_\text{max} / \sqrt{2} \). Alright? So you can always combine these equations with your other intensity equations and everything works out. Alright? So just be really careful that you know which one you're solving for, because a lot of questions will try to throw you off and ask for \( E_\text{RMS} \) or \( B_\text{max} \) or something like that. So just be really careful that you're solving the right thing here. Okay? So in this last part we're looking for is we're actually looking for \( B_\text{RMS} \). Alright? So I know that \( B_\text{RMS} \) is equal to \( B_\text{max} / \sqrt{2} \). The problem is I don't know what \( B_\text{max} \) is, but I can calculate it because I know that \( B_\text{max} \), remember, is always equal to, is always related to, \( E_\text{max} \) by the speed of light equation. So I can always calculate what \( B_\text{max} \) is because I have that \( E_\text{max} \) is equal to \( c B \). So therefore, your \( E_\text{max} \), which is your 10.98 divided by \( 3 \times 10^8 \) is equal to your \( B_\text{max} \), and that's going to equal, that's going to be \( 3.66 \times 10^{-8} \), and that's in Teslas. So, basically, what I can do here is I can say, I'm going to bring this all the way out here, and I'm going to say that \( B_\text{RMS} \) is equal to \( 3.66 \times 10^{-8} / \sqrt{2} \). And your final answer is going to be, that's going to be, \( 2.59 \times 10^{-8} \) in Teslas. Alright? So that's your final answer. Anyway, let me know if you have any questions, and I'll see you in the next one.

Example 1

Alright, folks. Let's get started with this practice problem here. So we're told that the average intensity of sunlight at the Earth's surface is about 1400 watts per square meter. Now remember that's I=1400. Now in this problem we're asked for 2 things; there are two parts. The first part asks us to calculate the amplitudes of the electric and magnetic fields for this wave. Now remember, whenever they're asking you for the amplitude, you should always think that's the maximum or the magnitude of the fields. So amplitude always means either bmax or emax, and in this case, they're actually asking us for both of them. So let's just go ahead and get started with that, right? So in part (a), they're asking us for emax and bmax. Now whenever you're relating emax and bmax, you have the intensity; you can always use this equation over here.

This I is equal to power times area, but it's also related to emax and bmax, and there are two separate equations for it. So we're just gonna go ahead and get started there. The first one we're gonna tackle is the emax one. So remember that the I is equal to power divided by area, but in this particular part, in part (a), we don't care about power or area, so this is not the expression we're looking for. But we do want to relate this to the emax. And remember this equation here says that I=12emax. Alright? So we have what the intensity is, and if we want to calculate Emax, then all we have to do is just rearrange this equation because remember c and ε00 are just constants. So what we're gonna do over here: we're gonna take this, we're gonna pull it over to the other side, and we're gonna say that I12cε00=emax. Now, by the way, you could move this one half over to the top if you want, but you could also leave it in the denominator, which I did.

Now the last thing we need to do here is this is gonna calculate what emax is, but we want Emax, and we just have to take the square roots. So this is just gonna be the square roots, and now we can just start plugging in some numbers here. So we have the square roots of 1400 divided by 12×3×108. And then ε00, remember, is just a constant, and all the constants are over here on the right side of this table, is 8.85×10-12. Alright? Now you don't have to put the units because the units will just cancel out, and you'll get something that is in Newtons per Coulomb. You can kinda just trust me on that. And this is gonna equal Emax, and so this is gonna be, if you actually plug this in, 1027 Newtons per Coulomb. Alright? So this is the final answer for your electric field.

Now that's one half of part (a) because we also want to calculate what is the magnetic field, bmax. Now bmax actually now that we've figured out Emax, you can actually solve it in one of two ways here. Alright? So I want to point out here that bmax can be solved using two ways. Now one of the ways you could do it is you could basically do exactly what we did for the electric field, but set the I equal to 12c&Divide;μ00bmax. Alright? So basically, we can do here is we could say that bmax=1/2c/μ00× I. So I is equal to 12c&Divide;μ00bmax, and you could basically go ahead and solve for this bmax using this method.

Or now that we've actually solved for the Emax, remember, you can always relate emax and bmax by using this equation over here, that emax=c×bmax. Alright. So or we can put that Emax=c×bmax. Now that we figured out what Emax is, you can actually just plug it into the equation. Alright? So, basically, whenever you figure out one of them, you can always figure out what the other one is by sort of taking this shortcut equation over here. Alright? So that's what we're gonna do because it's gonna be a little bit faster. So this is gonna be, emax÷c, which will equal bmax. And if you go ahead and plug this in, what you're gonna get is 1027÷3×108. And when you work this out, what you should get, is you should get 3.42×10 -6 in Teslas. Now what I want you to do is I wanted you to pause the video if you're just if you don't trust me, and you can actually work this out by using this equation over here, and you should get that bmax is equal to the same thing.

Alright? So let's go ahead and move on now to part (b). Alright? So that was part (a). Now part (b) asks us to find what? Now part (b) says, if the Earth-Sun distance is approximately 1.5×1011 meters, then what's the power output of the sun? So now we're looking for the average power output. Alright? So, basically, what that means here, is that we're looking for p. And, by the way, every time they say power the average power output, that p average really just means p. Alright? So there's really no difference between average power and power. It's all just the same p. Alright? So then how do we calculate this p over here? Now remember that power is always related to intensity via these two equations over here, I=p÷a. So we have what the intensity is. So we basically say that I=p÷a. Now the one that we need to do is we just need to figure out what kind of sort of source are we dealing with here. Are we dealing with a source that radiates outward radially or is it directional? Remember, if we're talking about the sun here, the sun is a spherical object and it radiates light everywhere. It's not like a laser beam or a flashlight. So, basically, what this means here is that your a is a sphere, which equals, 4πr2. Alright? So basically, what we can do here is we can say that the power is going to be equa

Example 2

Alright, folks. So let's take a look at this example problem here. We have a radio tower that's emitting some kind of radio signal with an average power output of 50,000 watts. In other words, that's our P or P average, which equals 50,000. This radio tower emits equally in a hemisphere above Earth's surface. Basically, what happens is if you could imagine that the ground is here, the signals can't penetrate through the ground, so it sort of radiates equally in a hemisphere, not a perfect sphere.

Now, we want to calculate the intensity that's detected by a satellite passing overhead at a height of 100 kilometers. From the radio tower up to this satellite, this is going to be my H, which equals 100 kilometers. So let's start with that first part, which is the intensity. To calculate the intensity, we just have this one big long expression to do this, which is just power divided by area. In this problem, we have no information about E_max, E_RMS, B_max, or B_RMS, or anything like that. So this whole problem can be solved just by using I = P/A. Right?

We have the power, and we need to calculate the area. The area of a hemisphere is really just equal to one-half of the area of an actual sphere, which is 1/2 of 4πr². So this means the area of a hemisphere that we're going to plug into our equation is actually 2πr². The radius is just the distance between the radio tower and the satellite, which is \( h = 100,000 \) meters. The intensity \( I \) is then calculated as:

50000 2 π 1 × 10 5 2

\( = 7.96 \times 10^{-7} \text{ watts per meter squared.} \)

Now, moving on to the second part, it asks us to calculate the power that's received by the satellite's circular radio antenna with a radius of 0.5 meters. The antenna captures signals with an intensity \( I \) and has an area of π(0.5)². Using \( P = I \times A \), the power that the antenna receives is:

7.96 × 10 -7 π 0.5 2

\( = 6.25 \times 10^{-7} \text{ watts.} \)

So, just to recap, in part A, we used the power of the radio tower and the area over which it broadcasts to figure out the intensity at a specific point. In part B, we reversed the equation to figure out the power received by a radio dish of a different size. Let me know if you have any questions in the comments, and I'll see you in the next video.

Here’s what students ask on this topic:

What is the intensity of electromagnetic waves?

Intensity (I) of electromagnetic waves is defined as the power (P) per unit area (A) through which the wave propagates. Mathematically, it is expressed as:

$I=\frac{P}{A}$

For electromagnetic waves, intensity can also be related to the electric field (E) and magnetic field (B) strengths. The equations are:

$I=\frac{1}{2}c\epsilon$₀Emax2

and

$I=\frac{1}{2}\frac{c}{\mu }$₀Bmax2

where $c$ is the speed of light, $\epsilon$₀ is the permittivity of free space, and $\mu$₀ is the permeability of free space.

How do you calculate the intensity of light at a certain distance from a source?

To calculate the intensity of light at a certain distance from a source, you need to know the power (P) of the light source and the area (A) over which the light spreads. For a source emitting light uniformly in all directions, the area is the surface area of a sphere:

$A=4\pi r^2$

where $r$ is the distance from the source. The intensity (I) is then given by:

$I=\frac{P}{A}=\frac{P}{4\pi r^2}$

This formula assumes no energy losses and uniform emission in all directions.

What is the relationship between intensity and electric field in electromagnetic waves?

The intensity (I) of electromagnetic waves is related to the maximum electric field (E_max) by the equation:

$I=\frac{1}{2}c\epsilon$₀Emax2

where $c$ is the speed of light in a vacuum, and $\epsilon$₀ is the permittivity of free space. This equation shows that the intensity is proportional to the square of the maximum electric field.

How do you find the RMS value of the magnetic field in an electromagnetic wave?

The RMS (root mean square) value of the magnetic field (B_RMS) in an electromagnetic wave can be found using the maximum magnetic field (B_max) with the relationship:

$B$_RMS=B_max√2

To find B_max, you can use the relationship between the electric field and magnetic field in an electromagnetic wave:

$E$_max=cB_max

where $c$ is the speed of light. Rearrange to solve for B_max:

$B$_max=E_maxc

Then, use the first equation to find B_RMS.

What are the units of intensity for electromagnetic waves?

The units of intensity (I) for electromagnetic waves are watts per square meter (W/m²). This unit is derived from the definition of intensity as power (P) per unit area (A):

$I=\frac{P}{A}$

Since power is measured in watts (W) and area in square meters (m²), the resulting unit for intensity is W/m².